Precalculus: Mathematics for Calculus, 7th Edition

by
Stewart, James; Redlin, Lothar; Watson, Saleem

Answer

$\frac{2}{(x+3)(x-1)}$

Work Step by Step

Join the two fractions $\frac{1}{x-1}+\frac{1}{x+3}$:
Find the LCD of the two fractions (i.e. $(x+3)(x-1)$) and adjust accordingly. This then becomes:
$\frac{1\times(x+3)}{(x+3)(x-1)}+\frac{1\times(x-1)}{(x+3)(x-1)}$
Combine the two fractions:
$\frac{(x+3)+(x-1)}{(x+3)(x-1)}$
Collect like terms:
$\frac{2x+2}{(x+3)(x-1)}$
So the fraction is now:
$=\frac{\frac{2x+2}{(x+3)(x-1)}}{x+1}$
Apply the fraction rule: $\frac{\frac{b}{c}}{a}=\frac{b}{c\times a}$ This then becomes:
$=\frac{2x+2}{(x+3)(x-1)(x+1)}$
Factor the numerator:
$=\frac{2(x+1)}{(x+3)(x-1)(x+1)}$
Cancel out the common factor:
$=\frac{2}{(x+3)(x-1)}$